In what algebraic structure does repeated addition equal multiplication?

I'm trying to figure out for which algebraic structure

$$\underbrace{a+a+\cdots+a}_{n \text{-times}} = a * n$$

is true.

Now I know the question 'Is all multiplication repeated addition?' has been asked many times with the answer: NO because you cannot express non-integer (such as fractions or complex numbers) multiples as repeated addition. However I'm pretty sure that the reverse is true; that 'Repeated addition is always multiplication'

So my first thought was that Rings would be the appropriate algebraic structure for this, seeing that they have both addition and multiplication. However, the definition of a ring does not mention this property.

So I was thinking about this property and it seems like it holds for many rings, including the following:

• Integers
• Rationals
• Reals
• Complex numbers
• $m\times m$ Matrix Ring
• Polynomials where multiplication is scaling by a number

But... Then I ran into the Boolean ring where $\lor$ is the addition in the ring, and $\land$ is the multiplication. So...

$$???\,\,\underbrace{a\lor a\lor\cdots\lor a}_{n \text{-times}} = a \land n \,\,???$$

Now the problem is the type of the entity is totally different (true/false values). This doesn't even make sense; or does it? If this isn't true, then I'm not sure where that leaves me then, since it would imply that this property doesn't hold for rings in general. But then what does it hold for?

Any insight would be greatly apprecitated. :)

• Any ring is automatically (and uniquely) also a $\mathbb{Z}$-algebra, and it is then true that $a + \cdots + a = n \cdot a$ using this $\mathbb{Z}$-algebra structure for the right hand side. (In the case of Boolean rings, it just happens that $2 \cdot x = 0$ for all $x$ in the ring, and more generally $n \cdot x = 0$ whenever $n$ is even.) – Daniel Schepler Mar 15 '18 at 1:06
• Any structure that has a distributive law $a(b+c) = ab + ac$ upon any element that is a finite sum of additive identities, this will hold as if $m= 1+1+1 ... + 1$ then $m*n = (1+1+1.... +1)n = n+n+n+n+....+n$. but if neither are the finite sum of 1s (i.e. an integer) then this simply isn't meaningfl – fleablood Mar 15 '18 at 1:24
• @DanielSchepler I haven't done much with rings yet, but isn't $a \vee a \vee a$ the same as $a \wedge a \wedge a$ the same as just $a$ in boolean algebra? – Preston Millward Mar 15 '18 at 2:12
• @Preston the addition operation in a Boolean algebra is XOR, so $a+a=0$. This actually turns it into a vector space. – Matt Samuel Mar 15 '18 at 2:16
• @RyanStull: $\vee$ is not the addition operation of a ring. Either you meant to write the exclusive or (XOR) operation instead, or you instead meant to talk about a semiring. – Excluded and Offended Mar 15 '18 at 13:48

A very concise description of what's true here is that when:

• addition and multiplication are associative
• $1$ is the multiplicative identity

then for every positive integer $n$, you have

$$(\underbrace{1 + 1 + \ldots + 1}_{n \text{ times}}) x = \underbrace{x + x + \ldots + x}_{n \text{ times}} = x (\underbrace{1 + 1 + \ldots + 1}_{n \text{ times}})$$

• Hmm, why do addition and multiplication need to be associative? Couldn't this work without that? – Ryan Stull Mar 16 '18 at 5:50
• @RyanStull: I didn't attempt to make a minimal list of hypotheses; I'm sure you can do with less. – Excluded and Offended Mar 16 '18 at 6:34

Sort of! What you're talking about is the confluence of two things:

1. the notion of a power of an element in a group;
2. the integers acting on an abelian group.

Firstly, the first point says that for an element $a$ in a group $G$, there is the notion of the power of $a$, written as "$a^n$" when we are using multiplicative notation for the group operation and written as $n a$ when using additive notation for the group operation.

Secondly, the first point allows us to think about any Abelian group as a $\mathbb Z$-module, so that there's a bilinear multiplication $\mathbb Z\times M\to M$ satisfying $1m=m$ for all $m$, and therefore by bilinearity $m+m=(1+1)m$, etc.

Given any $R$-module $M$, this $\mathbb Z$ action is always captured by what $1\in R$ does to elements of $M$.

So I think the best answer to your question is an abelian group (aka a $\mathbb Z$-module.)

For any abelian group $M$, repeated addition of $m$ in $M$ is the same as the action of $n\in \mathbb Z$ on $m$.

We could clarify the highlighted statement for rings like this:

For a ring $R$ with identity and a unitary $R$-module $M$, repeated addition in $M$ is the same as multiplying by $n1$ where $1$ is the identity of $R$.

I don't agree with your statements about boolean rings though. $\vee$ and $\wedge$ are not the usual notations for the operations in a boolean ring: they are the usual notations for the lattice operation of a boolean algebra, though.

There is a link between the two: $a+b=(a\wedge\neg b)\vee (b\wedge\neg a)$ and $ab=a\wedge b$.

In that case $a+a+\ldots+a\in\{0,a\}$ depending on the parity of the number of $a$'s. This matches multiplication by $n$, because $n\equiv 0$ or $n\equiv 1$ based on the parity of $n$.